The reaction is a zero order reaction. A drop of a solution (volume =0.05 mL ) contains 6× 10- 7 mol of H+. If the rate of disappearance of H+ is 6.0× 105 mol L- 1s- 1, how long will it take for the H+ in the drop to disappear?

Question

The reaction is a zero order reaction. A drop of a solution (volume =0.05 mL ) contains 6× 10- 7 mol of H+. If the rate of disappearance of H+ is 6.0× 105 mol L- 1s- 1, how long will it take for the H+ in the drop to disappear? (A) 8.0× 10- 8s (B) 2.0× 10- 8s (C) 6.0× 10- 6s (D) 2.0× 10- 2s

Tardigrade Admin · Accepted Answer

[H+]=(6 × 10- 7 m o l /0.05 × 10- 3 L)=1.2× 10- 2M or r=(Δ x/Δ t) or Δ T=(Δ x/r)=(1.2 × 10- 2 M/6 × 105 M s- 1) ∴ t=2× 10- 8s.

Q. The reaction is a zero order reaction. A drop of a solution (volume $= 0.05 m L$ ) contains $6 \times 1 0^{- 7}$ mol of $H^{+} .$ If the rate of disappearance of $H^{+}$ is $6.0 \times 1 0^{5}$ mol $L^{- 1} s^{- 1},$ how long will it take for the $H^{+}$ in the drop to disappear?

$8.0 \times 1 0^{- 8} s$

$2.0 \times 1 0^{- 8} s$

$6.0 \times 1 0^{- 6} s$

$2.0 \times 1 0^{- 2} s$

$[H^{+}] = \frac{6 \times 1 0 ^{- 7} m o l}{0.05 \times 1 0 ^{- 3} L} = 1.2 \times 1 0^{- 2} M$
or $r = \frac{Δ x}{Δ t}$ or $Δ T = \frac{Δ x}{r} = \frac{1.2 \times 1 0 ^{- 2} M}{6 \times 1 0 ^{5} M s ^{- 1}}$
$∴ t = 2 \times 1 0^{- 8} s .$

Q. The reaction is a zero order reaction. A drop of a solution (volume =0.05mL ) contains 6×10−7 mol of H+. If the rate of disappearance of H+ is 6.0×105 mol L−1s−1, how long will it take for the H+ in the drop to disappear?

8.0×10−8s

2.0×10−8s

6.0×10−6s

2.0×10−2s